Question

Quentin works as a used car salesman, where he earns $8 per hour and an average commission of$260 per car sold. He wants to earn at least $1040 per week.Identify one point that IS a solution but is NOT reasonable in the real-world context. Justify yourresponse.

Answer 1

The point that is a solution but is not reasonable in the real-world context is (500, -11.4)

Identifying a not reasonable solution in the real-world context.

From the question, we have the following parameters that can be used in our computation:

The graph of the inequality expression

Also, we have

Earnings = $8 per hour

Average commission = $260 per car

So, the total earning is

Total = 8x + 260y

Where

x = Number of hours

y = Number of cars

He wants to earn at least $1040 per week.

So, we have

Total ≥ 1040

This means that

8x + 260y ≥ 1040

Divide through by 4

2x + 65y ≥ 260

Set x = 500

So, we have

2 * 500 + 65y ≥ 260

1000 + 65y ≥ 260

65y ≥ -740

Evaluate

y ≥ -11.4

We know that number of weeks cannot be negative

So, the value of y must start from 0

Hence, one point that is a solution but is not reasonable in the real-world context is (500, -11.4)

Answer 2

Inequalities

Let x = number of hours worked by Quentin.

y = number of cars sold by Quentin.

The total earnings are calculated as:

E = 8x + 280y

He wants to earn at least $1040 per week, thus:

8x + 280y ≥ 1040

Dividing by 8:

x + 35y ≥ 130

Solving for x:

x ≥ 130 - 35y

Supposse he sells not a single car (y = 0). Thus:

x ≥ 130

He needs to work a minimum of 130 hours per week.

One week has 7*24 = 168 hours.

It's possible to work that many hours a week, but it's not reasonable. He would only have 38 hours to sleep, rest, eat, etc.

One solution is y = 0, x = 130 but it's not reasonable in the context of the problem.

There are many other possible solutions. For example, if he sells y = 10 cars in the week (quite possible), then:

x ≥ 130 - 350

x ≥ -220

One solution could be x = -100, y = 10. but it's not possible to work for -10 hours.